For 2, you can see that modulo 2, \( f(a) \equiv f(1) , f(0) \equiv 1 \)( depending on whether a is odd or even). for all integral \( a \). But, if a integral root \( b \) did exist, that would imply that \( f(b) = 0 \equiv 0 \), which leads to a contradiction

Clearly, \(f_{n+1}^{\prime}(x)\) has no repeated roots and must have \(n\) distinct roots by our inductive hypothesis. Therefore, by Rolle's Theorem, we must have \(n+1\) distinct roots of \(f_{n+1}(x)\). And thus proved.

The first one is pretty simple. I don't like using LMV or Rolle's theorem, or at least their terms in anything that can be solved with a rough sketch. Although I don't use it, my method is pretty much related to it. If the function has repeated roots, then it's derivative also has a root at that point. If that's the case, then the \(x^n/n!\) factor in the function has no consequence in altering the value taken by the function. In that case, x must have a repeated root at 0. But substituting 0 in the place of x gives you 1. So 0 is not a root and the function doesn't have any repeated roots.